Arrangement of pie charts using matplotlib subplot

Question

I have 7 pi-charts (4 are listed below). I am trying to create a dashboard with 4 pie charts in first row and 3 pie charts in second row. Not sure where I am going wrong with the below code. Are there any other alternatives to achieve this? Any help would be appreciated.

from matplotlib import pyplot as PLT
fig = PLT.figure()
ax1 = fig.add_subplot(221)
line1 = plt.pie(df_14,colors=("g","r"))
plt.title('EventLogs')
ax1 = fig.add_subplot(223)
line2 = plt.pie(df_24,colors=("g","r"))
plt.title('InstalledApp')
ax1 = fig.add_subplot(222)
line3 = plt.pie(df_34,colors=("g","r"))
plt.title('Drive')
ax1 = fig.add_subplot(224)
line4 = plt.pie(df_44,colors=("g","r"))
plt.title('SQL Job')
ax1 = fig.add_subplot(321)
line5 = plt.pie(df_54,colors=("g","r"))
plt.title('Administrators')
ax2 = fig.add_subplot(212)
PLT.show()

bhargavkartik · Accepted Answer · 2021-08-03 05:58:26Z

8

A better method which I always use and is more intuitive, at-least for me, is to use subplot2grid....

fig = plt.figure(figsize=(18,10), dpi=1600)
#this line will produce a figure which has 2 row 
#and 4 columns 
#(0, 0) specifies the left upper coordinate of your plot
ax1 = plt.subplot2grid((2,4),(0,0))
plt.pie(df_14,colors=("g","r"))
plt.title('EventLogs')
#next one
ax1 = plt.subplot2grid((2, 4), (0, 1))
plt.pie(df_24,colors=("g","r"))
plt.title('InstalledApp')

And you can go on like this, and when you want to switch the row just write the coordinate as (1, 0)... which is second row-first column.

An example with 2 rows and 2 cols -

fig = plt.figure(figsize=(18,10), dpi=1600)
#2 rows 2 columns

#first row, first column
ax1 = plt.subplot2grid((2,2),(0,0))
plt.pie(df.a,colors=("g","r"))
plt.title('EventLogs')

#first row sec column
ax1 = plt.subplot2grid((2,2), (0, 1))
plt.pie(df.a,colors=("g","r"))
plt.title('EventLog_2')

#Second row first column
ax1 = plt.subplot2grid((2,2), (1, 0))
plt.pie(df.a,colors=("g","r"))
plt.title('InstalledApp')

#second row second column
ax1 = plt.subplot2grid((2,2), (1, 1))
plt.pie(df.a,colors=("g","r"))
plt.title('InstalledApp_2')

Hope this helps!

edited Aug 3, 2021 at 5:58

bhargavkartik

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answered Jul 18, 2016 at 13:52

hashcode55

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4 Comments

user3447653 Over a year ago

Thanks, it works perfectly. Is there a way to set a common legend for all the 7 plots? For example, red as "Not healthy", green as "Healthy"?

hashcode55 Over a year ago

You are looking for figlegend, but as each of the plots have their own axes, its not possible to do it with legend directly. Or why not just position your legend for one plot?

user3447653 Over a year ago

I tried this but not working as expected: lgd = plt.legend((line1), ('Helathy','Not Healthy'), loc='upper right')

Josh Friedlander Over a year ago

This answer, and subplot2grid, deserve much more attention. For such a basic task, subplots in Matplotlib are one of the most frustrating things I have to deal with. Bookmarking for future use!

T4gmaster · Accepted Answer · 2021-08-17 10:30:52Z

Use this if you want to create quicker arrangements of subplots

In addition to hashcode55's code:

When you want to avoid making multiple DataFrames, I recommend to assign integers to your feature-column and iterate through those. Make sure you make a dictionary for the features though. Here I am doing a plot with 4 columns and 2 rows.

fig = plt.figure(figsize=(25,10)) #,dpi=1600)
i= 0 #this is the feature I used
r,c = 0 ,0   #these are the rows(r) and columns(c)

for i in range(7):
  if c < 4:
    #weekday
    ax1 = plt.subplot2grid((2,4), (r, c))
    plt.pie(data[data.feature == i].something , labels = ..., autopct='%.0f%%')
    plt.title(feature[i])
    c +=1 #go one column to the left
    i+=1 #go to the next feature
else:
    c = 0 #reset column number as we exceeded 4 columns
    r = 1 #go into the second row
    ax1 = plt.subplot2grid((2,4), (r, c))
    plt.pie(data[data.feature == i].something , labels = ..., autopct='%.0f%%')
    plt.title(days[i])
    c +=1
    i+=1

plt.show()

This code will go on until the amount of features is exhausted.

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